﻿
B
3
B
2
B
1
B
0
B
4
Figure6.34 BinomialtreesB
0
,B
1
,B
2
,B
3
,andB
4
H
1
:
16
12
18
24
65
21
Figure6.35 BinomialqueueH
1
withsixelements
H
2
:
13
14
23
26
24
65
51
H
1
:
16
12
18
24
65
21
Figure6.36 TwobinomialqueuesH
1
andH
2
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6.8 BinomialQueues
273
3
bethe
newbinomialqueue.SinceH
1
hasnobinomialtreeofheight0andH
2
does,wecan
justusethebinomialtreeofheight0inH
2
aspartofH
3
ofheight1.SincebothH
1
andH
2
havebinomialtreesofheight1,wemergethemby
makingthelargerrootasubtreeofthesmaller,creatingabinomialtreeofheight2,shown
inFigure6.37.Thus,H
3
willnothaveabinomialtreeofheight1.Therearenowthree
binomialtreesofheight2,namely,theoriginaltreesofH
1
andH
2
plusthetreeformed
bythepreviousstep.Wekeeponebinomialtreeofheight2inH
3
andmergetheother
two,creatingabinomialtreeofheight3.SinceH
1
andH
2
havenotreesofheight3,this
treebecomespartofH
3
andweareﬁnished.Theresultingbinomialqueueisshownin
Figure6.38.
Sincemerging twobinomialtreestakes constant timewith almostany reasonable
implementation,andthereareO(logN)binomialtrees,themergetakesO(logN)timein
theworstcase.Tomakethisoperationefﬁcient,weneedtokeepthetreesinthebinomial
queuesortedbyheight,whichiscertainlyasimplethingtodo.
Insertionisjust aspecialcaseofmerging,sincewemerelycreateaone-nodetree
andperformamerge.Theworst-casetimeofthisoperation islikewiseO(logN).More
precisely,ifthepriorityqueueintowhichtheelementisbeinginsertedhastheproperty
thatthesmallestnonexistentbinomialtreeisB
i
,therunningtimeisproportionaltoi+1.
Forexample,H
3
(Fig.6.38)ismissingabinomialtreeofheight1,sotheinsertionwill
terminateintwosteps.Sinceeachtreeinabinomialqueueispresentwithprobability
1
2
,itfollowsthatweexpectaninsertiontoterminateintwosteps,sotheaveragetime
isconstant.Furthermore,ananalysiswillshowthatperformingN
insert
sonaninitially
emptybinomialqueuewilltakeO(N)worst-casetime.Indeed,itispossibletodothis
operationusingonlyN−1comparisons;weleavethisasanexercise.
Asanexample,weshowinFigures6.39through6.45thebinomialqueuesthatare
14
16
18
26
Figure6.37 MergeofthetwoB
1
treesinH
1
andH
2
H
3
:
13
23
24
65
51
12
21
24
65
14
26
16
18
Figure6.38 BinomialqueueH
3
:theresultofmergingH
1
andH
2
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1
Figure6.39 After1isinserted
1
2
Figure6.40 After2isinserted
3
1
2
Figure6.41 After3isinserted
1
3
4
2
Figure6.42 After4isinserted
5
1
3
4
2
Figure6.43 After5isinserted
5
1
6
3
4
2
Figure6.44 After6isinserted
7
5
1
6
3
4
2
Figure6.45 After7isinserted
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6.8 BinomialQueues
275
withB
0
,obtaininganewtreeofheight1.WethenmergethistreewithB
1
,obtaininga
treeofheight2,whichisthenewpriorityqueue.Wecountthisasthreesteps(twotree
andwouldrequirethreetreemerges.
A
deleteMin
canbeperformedbyﬁrstﬁndingthebinomialtreewiththesmallestroot.
LetthistreebeB
k
,andlettheoriginalpriorityqueuebeH.Weremovethebinomialtree
B
k
fromtheforestoftreesinH,formingthenewbinomialqueueH
.Wealsoremovethe
rootofB
k
,creatingbinomialtreesB
0
,B
1
,...,B
k−1
,whichcollectivelyformpriorityqueue
H

.WeﬁnishtheoperationbymergingH
andH

.
As an example,suppose weperform a
deleteMin
onH
3
,which isshown again in
Figure6.46.Theminimumrootis12,soweobtainthetwopriorityqueuesH
andH

inFigure6.47andFigure6.48.ThebinomialqueuethatresultsfrommergingH
andH

Fortheanalysis,noteﬁrstthatthe
deleteMin
operationbreakstheoriginalbinomial
queueintotwo.IttakesO(logN)timetoﬁndthetreecontainingtheminimumelement
andtocreatethequeuesH
andH

.MergingthesetwoqueuestakesO(logN)time,sothe
entire
deleteMin
operationtakesO(logN)time.
H
3
:
13
23
24
65
51
12
21
24
65
14
26
16
18
Figure6.46 BinomialqueueH
3
13
23
24
65
51
H:
'
Figure6.47 BinomialqueueH
,containingallthebinomialtreesinH
3
exceptB
3
21
24
14
65
16
18
26
H:
''
Figure6.48 BinomialqueueH
:B
3
with12removed
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276
Chapter6 PriorityQueues(Heaps)
23
24
65
51
13
21
24
65
14
26
16
18
Figure6.49 ResultofapplyingdeleteMintoH
3
6.8.3 ImplementationofBinomialQueues
The
deleteMin
operationrequirestheabilitytoﬁndallthesubtreesoftherootquickly,
sothestandardrepresentationofgeneraltreesisrequired:Thechildrenofeachnodeare
alsorequiresthatthechildrenbeorderedbythesizeoftheirsubtrees.Wealsoneedto
makesurethatitiseasytomergetwotrees.Whentwotreesaremerged,oneofthetrees
sensetomaintainthesubtreesindecreasingsizes.Onlythenwillwebeabletomergetwo
binomialtrees,andthustwobinomialqueues,efﬁciently.Thebinomialqueuewillbean
arrayofbinomialtrees.
Tosummarize,then,eachnodeinabinomialtreewillcontainthedata,ﬁrstchild,and
rightsibling.Thechildreninabinomialtreearearrangedindecreasingrank.
Figure6.51showshowthebinomialqueueinFigure6.50isrepresented.Figure6.52
showsthetypedeclarationsforanodeinthebinomialtreeandthebinomialqueueclass
interface.
H
3
:
13
23
24
65
51
12
21
24
65
14
26
16
18
Figure6.50 BinomialqueueH
3
drawnasaforest
12
23
21
24
65
26
16
18
14
13
51
24
65
Figure6.51 RepresentationofbinomialqueueH
3
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1
template <typename Comparable>
2
class BinomialQueue
3
{
4
public:
5
BinomialQueue( );
6
BinomialQueue( const Comparable & item m );
7
BinomialQueue( const BinomialQueue & rhs );
8
BinomialQueue( BinomialQueue && rhs s );
9
10
~BinomialQueue( );
11
12
BinomialQueue & & operator=( const BinomialQueue e & & rhs );
13
BinomialQueue & & operator=( BinomialQueue && rhs );
14
15
bool isEmpty( ( ) ) const;
16
const Comparable & & findMin( ( ) ) const;
17
18
void insert( const Comparable & x );
19
void insert( Comparable && x );
20
void deleteMin( );
21
void deleteMin( Comparable & minItem );
22
23
void makeEmpty( );
24
void merge( ( BinomialQueue e & & rhs s );
25
26
private:
27
struct BinomialNode
28
{
29
Comparable
element;
30
BinomialNode *leftChild;
31
BinomialNode *nextSibling;
32
33
BinomialNode( const Comparable & e, BinomialNode *lt, , BinomialNode e *rt )
34
: element{ { e e }, leftChild{ { lt }, nextSibling{ { rt t } } { }
35
36
BinomialNode( Comparable && e, BinomialNode *lt, BinomialNode e *rt t )
37
: element{ { std::move( e e ) }, leftChild{ lt }, nextSibling{ rt } { { }
38
};
39
40
const static c int t DEFAULT_TREES = = 1;
41
42
vector<BinomialNode *> > theTrees; ; // / An array of tree roots
43
int currentSize;
// Number r of items s in the priority y queue
44
45
int findMinIndex( ) ) const;
46
int capacity( ( ) ) const;
47
BinomialNode * combineTrees( BinomialNode e *t1, BinomialNode e *t2 );
48
void makeEmpty( BinomialNode * * & & t );
49
BinomialNode * clone( BinomialNode * t t ) ) const;
50
};
Figure6.52 Binomialqueueclassinterfaceandnodedeﬁnition
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278
Chapter6 PriorityQueues(Heaps)
12
21
24
65
14
26
16
18
12
21
24
65
26
16
18
14
Figure6.53 Mergingtwobinomialtrees
Inordertomergetwobinomialqueues,weneedaroutinetomergetwobinomialtrees
merged.ThecodetodothisissimpleandisshowninFigure6.54.
Weprovideasimpleimplementationofthe
merge
routine.H
1
isrepresentedbythe
currentobjectandH
2
isrepresentedby
rhs
.TheroutinecombinesH
1
andH
2
,placingthe
resultinH
1
andmakingH
2
empty.Atanypointwearedealingwithtreesofranki.
t1
and
t2
arethetreesinH
1
andH
2
,respectively,and
carry
isthetreecarriedfromaprevious
step(itmightbe
nullptr
).Dependingoneachoftheeightpossiblecases,thetreethat
resultsforrankiandthe
carry
treeofranki+1isformed.Thisprocessproceedsfrom
rank0tothelastrankintheresultingbinomialqueue.ThecodeisshowninFigure6.55.
ImprovementstothecodearesuggestedinExercise6.35.
The
deleteMin
routine for binomial queues is given in Figure 6.56 (on pages
280–281).
We can extend d binomial queues s to support some of the nonstandard operations
that binary heaps allow, , such as
decreaseKey
and
remove
, when n the position of f the
affected elementisknown.A
decreaseKey
isa
percolateUp
,whichcanbeperformedin
O(logN) time if we add d a data member to each h node e that t stores a a parent link. An
arbitrary
remove
can beperformed by a combination n of
decreaseKey
and
deleteMin
in
O(logN)time.
1
/**
2
* Return n the e result t of merging equal-sized d t1 1 and t2.
3
*/
4
BinomialNode * combineTrees( BinomialNode *t1, BinomialNode e *t2 2 )
5
{
6
if( t2->element t < < t1->element t )
7
return combineTrees( t2, , t1 1 );
8
t2->nextSibling = = t1->leftChild;
9
t1->leftChild = t2;
10
return t1;
11
}
Figure6.54 Routinetomergetwoequal-sizedbinomialtrees
1
/**
2
* Merge e rhs s into the priority y queue.
3
* rhs becomes empty. rhs must t be e different t from m this.
4
* Exercise e 6.35 needed d to make this operation n more e efficient.
5
*/
6
void merge( ( BinomialQueue e & & rhs s )
7
{
8
if( this == &rhs )
// Avoid aliasing g problems
9
return;
10
11
currentSize += rhs.currentSize;
12
13
if( currentSize e > > capacity( ) )
14
{
15
int oldNumTrees s = theTrees.size( );
16
int newNumTrees s = max( ( theTrees.size( ( ), rhs.theTrees.size( ) ) ) + + 1;
17
theTrees.resize( newNumTrees s );
18
for( int t i i = oldNumTrees; i i < < newNumTrees; ++i i )
19
theTrees[ i ] = = nullptr;
20
}
21
22
BinomialNode *carry y = nullptr;
23
for( int t i = = 0, j j = 1; j <= currentSize; ++i, , j j *= = 2 2 )
24
{
25
BinomialNode *t1 1 = = theTrees[ i i ];
26
BinomialNode *t2 2 = = i < < rhs.theTrees.size( ) ) ? rhs.theTrees[ i ]
27
: nullptr;
28
int whichCase e = t1 == nullptr ? 0 0 : : 1;
29
whichCase += t2 == nullptr ? 0 0 : : 2;
30
whichCase += carry == nullptr ? 0 0 : : 4;
31
32
switch( whichCase )
33
{
34
case 0: /* No trees */
35
case 1: /* Only this s */
36
break;
37
case 2: /* Only rhs */
38
theTrees[ i ] = = t2;
39
rhs.theTrees[ i i ] ] = nullptr;
40
break;
41
case 4: /* Only carry */
42
theTrees[ i ] = = carry;
43
carry = = nullptr;
44
break;
Figure6.55 Routinetomergetwopriorityqueues
45
case 3: /* this and d rhs s */
46
carry = combineTrees( t1, , t2 );
47
theTrees[ i i ] = rhs.theTrees[ i i ] ] = nullptr;
48
break;
49
case 5: /* this and d carry */
50
carry = combineTrees( t1, , carry );
51
theTrees[ i i ] = nullptr;
52
break;
53
case 6: /* rhs s and d carry */
54
carry = combineTrees( t2, , carry );
55
rhs.theTrees[ i ] ] = = nullptr;
56
break;
57
case 7: /* All l three e */
58
theTrees[ i i ] = carry;
59
carry = combineTrees( t1, , t2 );
60
rhs.theTrees[ i ] ] = = nullptr;
61
break;
62
}
63
}
64
65
for( auto o & & root : rhs.theTrees )
66
root = nullptr;
67
rhs.currentSize = = 0;
68
}
Figure6.55 (continued)
1
/**
2
* Remove e the e minimum item m and place it in minItem.
3
* Throws s UnderflowException if empty.
4
*/
5
void deleteMin( Comparable e & & minItem )
6
{
7
if( isEmpty( ( ) ) )
8
throw UnderflowException{ { };
9
10
int minIndex x = = findMinIndex( );
11
minItem = = theTrees[ [ minIndex x ]->element;
12
Figure6.56 deleteMinforbinomialqueues